

A001105


a(n) = 2*n^2.


219



0, 2, 8, 18, 32, 50, 72, 98, 128, 162, 200, 242, 288, 338, 392, 450, 512, 578, 648, 722, 800, 882, 968, 1058, 1152, 1250, 1352, 1458, 1568, 1682, 1800, 1922, 2048, 2178, 2312, 2450, 2592, 2738, 2888, 3042, 3200, 3362, 3528, 3698, 3872, 4050, 4232, 4418
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OFFSET

0,2


COMMENTS

"If each period in the periodic system ends in a rare gas ..., the number of elements in a period can be found from the ordinal number n of the period by the formula: L = ((2n+3+(1)^n)^2)/8..."  Nature, Jun 09 1951; Nature 411 (Jun 07 2001), p. 648. This produces the present sequence doubled up.
Let z(1) = I; (I^2 = 1), z(k+1) = 1/(z(k)+2I); then a(n) = (1)*Imag(z(n+1))/real(z(n+1)).  Benoit Cloitre, Aug 06 2002
Maximum number of electrons in an atomic shell with total quantum number n. Partial sums of A016825.  Jeremy Gardiner, Dec 19 2004
Arithmetic mean of triangular numbers in pairs: (1+3)/2, (6+10)/2,(15+21)/2, ... .  Amarnath Murthy, Aug 05 2005
These numbers form a pattern on the Ulam spiral similar to that of the triangular numbers.  G. Roda, Oct 20 2010
Integral areas of isosceles right triangles with rational legs (legs are 2n and triangles are nondegenerate for n > 0).  Rick L. Shepherd, Sep 29 2009
Number of stars when distributed as in the U.S.A. flag: n rows with n+1 stars and, between each pair of these, one row with n stars (i.e., n1 of these), i.e., n*(n+1)+(n1)*n = 2*n^2 = A001105(n).  César Eliud Lozada, Sep 17 2012
Apparently the number of Dyck paths with semilength n+3 and an odd number of peaks and the central peak has height n3.  David Scambler, Apr 29 2013
Sum of the partition parts of 2n into exactly two parts.  Wesley Ivan Hurt, Jun 01 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective odd leg a (A180620); sequence gives values ca, sorted with duplicates removed.  K. G. Stier, Nov 04 2013
Number of roots in the root systems of type B_n and C_n (for n>1).  Tom Edgar, Nov 05 2013
This sequence appears also as the first and second member of the quartet [a(n), a(n), p(n), p(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p(n) = A046092(n). See an Oct 15 2014 comment on A147973 where also a reference is given.  Wolfdieter Lang, Oct 16 2014
a(n) are the only integers, m, where (A000005(m) + A000203(m)) = (number of divisors of m + sum of divisors of m) is an odd number.  Richard R. Forberg, Jan 09 2015
a(n) represents the first term in a sum of consecutive integers running to a(n+1)1 that equals (2n+1)^3.  Patrick J. McNab, Dec 24 2016
Also the number of 3cycles in the (n+4)triangular honeycomb obtuse knight graph.  Eric W. Weisstein, Jul 29 2017
Also the Wiener index of the ncocktail party graph for n > 1.  Eric W. Weisstein, Sep 07 2017
Numbers represented as the palindrome 242 in number base B including B=2(binary), 3(ternary) and 4: 242(2)=18, 242(3)=32, 242(4)=50, ... 242(9)=200, 242(10)=242, ...  Ron Knott, Nov 14 2017
a(n) is the square of the hypotenuse of an isosceles right triangle whose sides are equal to n.  Thomas M. Green, Aug 20 2019
Apart from 0, integers such that the number of even divisors (A183063) is odd.
Proof: every n = 2^q * (2k+1), q, k >= 0, then 2*n^2 = 2^(2q+1) * (2k+1)^2; now, gcd(2, 2k+1) = 1, tau(2^(2q+1)) = 2q+2 and tau((2k+1)^2) = 2u+1 because (2k+1)^2 is square, so, tau(2*n^2) = (2q+2) * (2u+1).
The 2q+2 divisors of 2^(2q+1) are {1, 2, 2^2, 2^3, ..., 2^(2q+1)}, so 2^(2q+1) has 2q+1 even divisors {2^1, 2^2, 2^3, ..., 2^(2q+1)}.
Conclusion, these 2q+1 even divisors create with the 2u+1 odd divisors of (2k+1)^2 exactly (2q+1)*(2u+1) even divisors of 2*n^2, and (2q+1)*(2u+1) is odd. (End)
a(n) with n>0 are the numbers with period length 2 for Bulgarian and Mancala solitaire.  Paul Weisenhorn, Jan 29 2022
Number of points at L1 distance = 2 from any given point in Z^n.  Shel Kaphan, Feb 25 2023


REFERENCES

Arthur Beiser, Concepts of Modern Physics, 2nd Ed., McGrawHill, 1973.
Martin Gardner, The Colossal Book of Mathematics, Classic Puzzles, Paradoxes and Problems, Chapter 2 entitled "The Calculus of Finite Differences," W. W. Norton and Company, New York, 2001, pages 1213.
L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 44.
Alain M. Robert, A Course in padic Analysis, SpringerVerlag, 2000, p. 213.


LINKS



FORMULA

a(n) = (1)^(n+1) * A053120(2*n, 2).
G.f.: 2*x*(1+x)/(1x)^3.
Sum_{n>=1} 1/a(n) = Pi^2/12 [Jolley eq. 319].  Gary W. Adamson, Dec 21 2006
a(n) = A002378(n1) + A002378(n).  Joerg M. Schuetze (joerg(AT)cyberheim.de), Mar 08 2010 [Corrected by Klaus Purath, Jun 18 2020]
For n > 0, a(n) = 1/coefficient of x^2 in the Maclaurin expansion of 1/(cos(x)+n1).  Francesco Daddi, Aug 04 2011
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+jn+4)/3).  Wesley Ivan Hurt, Mar 12 2015
Product_{n>=1} (1 + 1/a(n)) = sqrt(2)*sinh(Pi/sqrt(2))/Pi.
Product_{n>=1} (1  1/a(n)) = sqrt(2)*sin(Pi/sqrt(2))/Pi. (End)


EXAMPLE

a(3) = 18; since 2(3) = 6 has 3 partitions with exactly two parts: (5,1), (4,2), (3,3). Adding all the parts, we get: 1 + 2 + 3 + 3 + 4 + 5 = 18.  Wesley Ivan Hurt, Jun 01 2013


MAPLE



MATHEMATICA

LinearRecurrence[{3, 3, 1}, {2, 8, 18}, {0, 20}] (* Eric W. Weisstein, Jul 28 2017 *)


PROG

(Haskell)


CROSSREFS

Cf. numbers of the form n*(n*kk+4))/2 listed in A226488.
Integers such that: this sequence (the number of even divisors is odd), A028982 (the number of odd divisors is odd), A028983 (the number of odd divisors is even), A183300 (the number of even divisors is even).


KEYWORD

nonn,easy


AUTHOR

Bernd.Walter(AT)frankfurt.netsurf.de


STATUS

approved



